Question

Suppose a student mixes 4.00 mL of 2.00 # 10-3 M Fe(NO3 ) 3 with 5.00 mL of 2.00 # 10-3 M KSCN and 1.00 mL of water. The student then determines the [FeNCS2+] at equilibrium to be 8.75 # 10-5 M. Find the equilibrium constant for the following reaction. Show all your calculations for each step. Fe3+ (aq) + SCN- (aq) FeNCS2+ (aq) Step 1. Calculate the initial number of moles of Fe3+ and SCN- (use Equation 12). moles of Fe3+ _____________ moles of SCN- ______________ Step 2. How many moles of FeNCS2+ are present at equilibrium? What is the volume of the equilibrium mixture? ________________ mL moles of FeNCS2+ ____________ How many moles of Fe3+ and SCN- are consumed to produce the FeNCS2+? moles of Fe3+ _____________ moles of SCN- ______________

Answer 1

`\mathrm{Fe} 3+(\mathrm{ag})+\mathrm{SCN}-(\mathrm{aq})` <---->
`\mathrm{FeSCN} 2+(\mathrm{aq})`

Step-by-step explanation:

Use the concentration and volume values they've given you to find the moles of Fe(NO3)3 and KSCN that were initially present (before they were mixed)

moles = concentration in M x volume in L

n(Fe(NO3)3) =
`\left(2.00 * 10^(\wedge)-3\right) *(5 / 1000)=0.00001 \mathrm{mol}`
`=10^(\wedge)-5 \text { mol }`

n(KSCN) =
`\left(2.00 * 10^(\wedge)-3\right) *(4 / 1000)=0.000008 \mathrm{mol}`
`=8 * 10^(\wedge)-6 \text { mol }`

n(Fe(NO3)3) = n(Fe3+) =
`10^(\wedge)-5 \text { mol }`

`n(KSCN) = n(SCN-) =`
`8 * 10^(\wedge)-6 \text { mol }`

so n(Fe3+) initially present =
`10^(\wedge)-5 \text { mol }`

n(SCN-) initially present =
`8 * 10^(\wedge)-6 \text { mol }`